题目内容
已知AB∥CD,AD∥BC,点E在AB的延长线上,若∠ACD=30°,∠CAD=31°,则∠CBE=______,∠ADC+∠DCB+∠ABC+∠BAD=______.
∵AB∥CD,
∴∠CAB=∠ACD=30°
∴∠DAB=∠CAD+∠CAB=30°+31°=61°
∵AD∥BC
∴∠CBE=∠DAB=61°.
∵AD∥BC,
∴∠ADC+∠DCB=180°,∠ABC+∠BAD=180°
∴∠ADC+∠DCB+∠ABC+∠BAD=180°+180°=360°.
故答案是:61°,360°.
∴∠CAB=∠ACD=30°
∴∠DAB=∠CAD+∠CAB=30°+31°=61°
∵AD∥BC
∴∠CBE=∠DAB=61°.
∵AD∥BC,
∴∠ADC+∠DCB=180°,∠ABC+∠BAD=180°
∴∠ADC+∠DCB+∠ABC+∠BAD=180°+180°=360°.
故答案是:61°,360°.
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